Koch's Snowflake Assignment
Begin with an equilateral triangle. Then remove the inner third of each side of the triangle and construct an equilateral triangle where the segment was removed. By repeating this process multiple times, we create an image known as Koch's Snowflake.
The link below will direct you to the Logo website where you can enter the commands to create Koch's Snowflake.
Logo

Level	Logo Commands	Image of Koch's Snowflake	Perimeter of n-th Koch Snowflake
0	pen up gotoxy -120 90 pen on loop 3 forward 243 right 120 sleep 500 loopend		$\text{Perimeter} = L* S\\ L=\text{length of each line segment}\\ S=\text{number of line segments}\\\\\\ \text{Level 0}\\ L=243\\ S=3\\ \text{Perimeter}= 729$
1	pen up gotoxy -120 90 pen on loop 6 forward 81 left 60 forward 81 right 120 sleep 500 loopend		$\text{Perimeter} = L* S\\ L=\text{length of each line segment}\\ S=\text{number of line segments}\\\\\\ \text{Level 1}\\ L= 81\\ S= 3*4= 12\\ \text{Perimeter}= 972$
2	pen up gotoxy -120 90 pen on loop 6 forward 27 left 60 forward 27 right 120 forward 27 left 60 forward 27 left 60 forward 27 left 60 forward 27 right 120 forward 27 left 60 forward 27 right 120 loopend		$\text{Perimeter} = L* S\\ L=\text{length of each line segment}\\ S=\text{number of line segments}\\\\\\ \text{Level 2}\\ L= 27\\ S=3*4*4=3*4^{2} = 48\\ \text{Perimeter}=1296$
n			$\text{Perimeter of n-th Koch Snowflake}\\ =L*S\\ =x(\frac{1}{3})^{n}*(3*4^{n})\\\\ \text{L=length of each line segment}\\ \text{x=the original length of each line segment}\\ \frac{1}{3}^{n}=\text{factor by which each length decreases}\\\\ \text{S=the number of line segments}\\ \text{3=the number of sides in a triangle}\\ 4^{n}=\text{the factor by which the S increases}$

What is the perimeter of the "last" level of Koch's snowflake?

If we take our perimeter equation from above with x= 243 and n=100th level, we have the following:
$243(\frac{1}{3})^{100}*(3*4^{100})= 2.27*10^{15}$
If we have n continually increasing, the perimeter will also increase substantially.
Let's try infinity as our n value.
Taking the limit of our perimeter then gives us:
$\lim_{n\to \infty}({243(\frac{1}{3})^{\infty}*(3*4^{\infty}))= \infty$
Therefore the perimeter of the "last" level of Koch's snowflake is infinite.

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